# Finding lines that are not parallel to one another

Given the problem:

Let u = (3,3,2) and v = (-1,0,1). Which one of the following lines is not parallel to the other three?

a) x = 2u + tv

b) x = u + 2tv

c) x = tu + v

d) x = 3tv

I must be doing it wrong as I get (5, 6, 5) for (a) from 2(3, 3, 2) + t(-1, 0, 1), and (1, 3, 4) for (b) from (3, 3, 2) + 2t(-1, 0, 1) for example when I try to substitute u and v into the line equations of (a) and (b). I tried solving for c here, (5c, 6c, 5c) = (1, 3, 4) to test if (a) and (b) are parallel, but they aren't, which isn't right as the answer is supposed to be option (c), where x = tu + v. What is the correct method of finding a line is not parallel to others?

• You found one point on each of two lines. There is no way to tell whether lines are parallel by looking at just one point on each line. For a moment, forget about this particular problem and think how you can know whether lines are parallel when you are given them in vector form. – David K Sep 11 at 13:34
• Define a line that you know intersects one of the lines. Then determine which of the others intersect that line at the same angles as the first, and so on until one intersects at different angles, and that is your line that is non parallel to the others en.wikipedia.org/wiki/File:Parallel_transversal.svg – Adam Sep 11 at 13:35
• @Adam These lines are in three-dimensional space. – David K Sep 11 at 13:42
• I guess so, but the principle still applies albeit a bit more leg work no? Surely I mean they aren't even three dimensional objects – Adam Sep 11 at 14:07
• if line a) is perpendicular to our reference transversal, is b) and d) also going to be perpendicular to that line if they are parallel to a)? – Adam Sep 11 at 14:10

The lines a), b), and d) have the direction given by vector $$\mathbf v$$, and therefore are parallel to each other. But line c) has the direction given by vector $$\mathbf u$$, which is not parallel to $$\mathbf v$$. So, yes, the answer is c).